MYSQL PHP API-在另一個表的行中顯示和獲取多行

Question

我正在為我的移動應用程序創建一個API。 我正在使用PHP MYSQL和Slim框架（與該問題無關）進行開發。

我試圖從我的mysql數據庫中提取多個“ venues”，並為每個“ venue”檢索多個“ venue_images”。 數據庫：

venues        venue_images
------        ------------
id PK         image_venue_id FK (to venues)
venue_name    image_path
active

然后，我需要以這種格式輸出數據：

{
  "completed_in":0.01068,
  "returned":10,
  "results":[
    {
      "venue_id":"1",
      "venue_name":"NameHere",
      "images": [
         {
           "image_path":"http://www.pathhere.com"
         },
         {
           "image_path":"http://www.pathhere2.com"
         }
      ]
    }
  ]
}

因此，基本上，每個場所對圖像進行多次迭代。

我當前的代碼是：

$sql = "
    SELECT
         venues.id, venues.venue_name, venues.active,
         venue_images.image_venue_id, venue_images.image_path
    FROM
         venues
    LEFT JOIN 
         venue_images ON venue_images.image_venue_id = venues.id
    WHERE
         venues.active = 1
    LIMIT 0, 10
    ";

    $data = ORM::for_table('venues')->raw_query($sql, array())->find_many();
         if($data) {
        foreach ($data as $post) {
            $results[] = array (
                 'venue_id' => $post->id,
                 'venue_name' => $post->venue_name,
                 'images' => $post->image_path
            );
        }

        //Build full json
        $time = round((microTimer() - START_TIME), 5);
        $result = array(
             'completed_in' => $time,
             'returned' => count($results),
             'results' => $results
        );
        //Print JSON
        echo indent(stripslashes(json_encode($result)));
    } else {
         echo "Nothing found";
    }

我當前的代碼有效，但是會產生以下結果：

{
  "completed_in":0.01068,
  "returned":10,
  "results":[
    {
      "venue_id":"1",
      "venue_name":"The Bunker",
      "images":"https://s3.amazonaws.com/barholla/venues/1352383950-qPXNShGR6ikoafj_n.jpg"
    },
    {
      "venue_id":"1",
      "venue_name":"The Bunker",
      "images":"https://s3.amazonaws.com/barholla/venues/1352384236-RUfkGAWsCfAVdPm_n.jpg"
    }
]
}

“地堡”有兩個圖片。 而不是將圖像存儲在場所數組中，而是創建帶有第二張圖像的“ The Bunker”重復行。 就像我之前說過的，我需要在每個場所內迭代多個圖像。 任何幫助將非常感激！ 謝謝！

Answer 1

您想使用GROUP_CONCAT

像這樣的東西（可能不是100％准確：））

$sql = "
    SELECT
         v.id, v.venue_name, v.active,
         GROUP_CONCAT(i.image_path) as venue_image_string
    FROM
         venues v
    LEFT JOIN 
         venue_images i ON i.image_venue_id = v.id
    WHERE
         v.active = 1
    GROUP BY i.image_venue_id
    LIMIT 0, 10
";

您可能需要擺弄一些東西，但應該把自己放在正確的軌道上（注意：以CSV格式提供venue_image_string）

Answer 2

為什么不能使用多個查詢代替..？ 它更快，更簡單..！

$sql = "SELECT venues.id, venues.venue_name, venues.active FROM venues WHERE venues.active = 1 LIMIT 0, 10";

    $data = ORM::for_table('venues')->raw_query($sql, array())->find_many();
         if($data) {
        foreach ($data as $post) {
            $results[] = array ();

$sql = "SELECT image_path FROM venue_images WHERE image_venue_id = $post->id"; 
$images = ORM::for_table('venue_images')->raw_query($sql, array())->find_many();

 $results[] = array (
                 'venue_id' => $post->id,
                 'venue_name' => $post->venue_name, 
                 'images' => $images);
        }

        //Build full json
        $time = round((microTimer() - START_TIME), 5);
        $result = array(
             'completed_in' => $time,
             'returned' => count($results),
             'results' => $results
        );
        //Print JSON
        echo indent(stripslashes(json_encode($result)));
    } else {
         echo "Nothing found";
    }